JEE Main 2012 — Center of Mass Momentum and Collision Question with Solution
JEE Main 2012 (12 May Online)
Question
Two bodies and of mass and respectively are placed on a smooth floor. They are connected by a spring of negligible mass. third body of mass is placed on the floor. The body moves with a velocity along the line joining and and collides elastically with . At a certain time after the collision it is found that the instantaneous velocities of and are same and the compression of the spring is . The spring constant will be
Choose an option
Show full solutionCorrect option: D
Correct answer
D
Step-by-step explanation
Initial momentum of the system block . After striking with , the block comes to rest and now both block A and B moves with velocity when compression in spring is . By the law of conservation of linear momentum By the law of conservation of energy K.E. of block K.E. of system + P.E. of system
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Center of Mass Momentum and Collision chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2012, covering the Center of Mass Momentum and Collision chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.